Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int H[maxn], nx[maxn * 2], to[maxn * 2], numofE;
vector<int> G[maxn];
int fa[maxn][20];
vector<int> F[maxn][20];
void init(int n) {
  for (int i = 0; i <= n; i++) G[i].clear(), H[i] = 0;
  numofE = 0;
}
void add(int a, int b) {
  numofE++;
  to[numofE] = b;
  nx[numofE] = H[a];
  H[a] = numofE;
}
int S[15];
void join(vector<int> P, vector<int> T, vector<int>& D, int a = 10) {
  int ge = 0, A1 = P.size(), A2 = T.size(), a1 = 0, a2 = 0;
  while (ge < a && (a1 < A1 || a2 < A2)) {
    if (a1 < A1 && (a2 >= A2 || P[a1] < T[a2])) {
      if (ge == 0 || P[a1] != S[ge - 1]) {
        S[ge++] = P[a1];
      }
      a1++;
    } else {
      if (ge == 0 || T[a2] != S[ge - 1]) {
        S[ge++] = T[a2];
      }
      a2++;
    }
  }
  D.clear();
  for (int i = 0; i < ge; i++) D.push_back(S[i]);
}
int dep[maxn];
void dfs(int cur, int per, int de) {
  sort(G[cur].begin(), G[cur].end());
  dep[cur] = de;
  join(G[cur], G[fa[cur][0]], F[cur][0]);
  for (int i = 1; i < 20; i++) {
    fa[cur][i] = fa[fa[cur][i - 1]][i - 1];
    join(F[cur][i - 1], F[fa[cur][i - 1]][i - 1], F[cur][i]);
  }
  for (int i = H[cur]; i; i = nx[i]) {
    int tt = to[i];
    if (tt == per) continue;
    fa[tt][0] = cur;
    dfs(tt, cur, de + 1);
  }
}
vector<int> ans;
void solve(int u, int v, int a) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 19; i >= 0; i--) {
    if (dep[fa[u][i]] >= dep[v]) {
      join(ans, F[u][i], ans, a);
      u = fa[u][i];
    }
    if (u == v) {
      join(ans, G[u], ans, a);
      return;
    }
  }
  for (int i = 19; i >= 0; i--) {
    if (fa[u][i] != fa[v][i]) {
      join(ans, F[u][i], ans, a);
      join(ans, F[v][i], ans, a);
      u = fa[u][i];
      v = fa[v][i];
    }
  }
  join(ans, F[u][0], ans, a);
  join(ans, F[v][0], ans, a);
}
int main() {
  int n, m, q;
  while (scanf("%d%d%d", &n, &m, &q) == 3) {
    for (int i = 1; i < n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      add(a, b);
      add(b, a);
    }
    for (int i = 1; i <= m; i++) {
      int a;
      scanf("%d", &a);
      G[a].push_back(i);
    }
    fa[1][0] = 1;
    dfs(1, 0, 0);
    for (int i = 0; i < q; i++) {
      int u, v, a;
      scanf("%d%d%d", &u, &v, &a);
      ans.clear();
      solve(u, v, a);
      a = min(a, int(ans.size()));
      printf("%d", a);
      for (int i = 0; i < a; i++) printf(" %d", ans[i]);
      printf("\n");
    }
  }
  return 0;
}